How To Calculate RAID 5 Storage?

How much RAID 5 storage do I need if according to our calculator we need a total of 440TB for surveillance storage?

How to calculate RAID 5 storage requirement?

Quintin,

For RAID5, add the size of one HDD per RAID Group plus around 10% for Windows overhead. A RAID Group typically consists of all of the data drives plus the parity drive(s) in one RAID chassis but there are occasions where a RAID chassis is split into multiple RAID Groups.

So, let's assume your RAID boxes have 16 drive slots and you populate them with 4TB drives in one RAID5 Group (something I wouldn't recommend - more on that later). 15 drives would be data and the 16th would be the parity drive. If the drives are 4TB, you would have 15x4=60x.9=54TB net to Windows from that RAID box.

But I would not recommend RAID5 for large-scale video storage, nor would I recommend putting 16 drives in one RAID Group; especially with ever-larger drives. I would recommend at least RAID6 because the more drives in a RAID Group and the larger the drives, the more likely you will have storage failures caused by UREs (unrecoverable read errors) during rebuilds, which will lose all of the data in that RAID Group. I would also state that the larger the hard drives, the smaller the number that should form a RAID Group.

Hard disks typically encounter read errors in 1x10-15 to 1x10-16 bits. 1TB is 8x10-12 bits so 60TB is almost 5x10-14, which is getting pretty close to the one in 10-15 HDD error rate. Hard disks fail, and that is a fact of life. So when the bad drive is replaced or if the system contains at least one global hot spare per chassis, the RAID will attempt to reconstruct the data on the missing drive from the parity data that is spread out on all of the drives. In that dense a system (16 drives), there is a strong likelihood that at least one chunk of parity data will be unreadable. If that happens, rebuild will stop and the entire RAID Group will be lost.

Also, the larger the RAID Group, the longer rebuilds take. Rebuild time is the most dangerous for a storage system since the system is reading and writing from parity data while writing the video data.

I typically recommend RAID6 systems with no more than "10+2" drives and one hot spare per chassis. Of course, that doesn't add up with 16-drive chassis so a fairly safe compromise might be 13 data drives, 2 parity drives and 1 hot spare drive per 16-bay chassis. With 4TB drives, that gives you 13x4=52x.9=47TB net storage per chassis.

Hey Carl.....maybe a little more detail? I think you forgot to provide a drawing!

How much RAID 5 storage do I need if according to our calculator we need a total of 440TB for surveillance storage?

Based on Carl's assumption of 4 TB drives and one RAID group per 16 drive chassis, here's how much total RAID storage that you will need using RAID 5 and 6, with and without hot spares.

• RAID 5 15 Data, 1 Parity, 0 Spare = 8 groups=8 parity + 0 spares=32TB=472TB
• RAID 5, 14 Data, 1 Parity, 1 Spare = 8 groups=8 parity + 8 spares=64TB=504TB
• RAID 6, 14 Data, 2 Parity, 0 Spare = 8 groups=16 parity + 0 spares=64TB=504TB
• RAID 6, 13 Data, 2 Parity, 1 Spare = 9 groups=18 parity + 9 spares=108TB=548TB

Agree with Carl that running RAID 5 is not advisable with the advent of large capacity drives, I included just so you could see the delta. As for spares, they are optional. They let the rebuild process start immediately, which can be important if there is no IT support. Maybe Carl would offer an opinion on whether he sees hot spares in every chassis as critical...

'A',

I certainly wouldn't put a system in without at least one global spare per RAID chassis or one spare per RAID Group, but that's me. IMO, it is an absolute necessity with RAID5 but since RAID6 can tolerate two failed drives without losing data, I've seen it handled both ways. It depends on how fast a failed drive will get replaced.

Our system uses 9+2 RAID6 running 3TB drives with a hot spare for each RAID Group in Dell MD3xxx 60-bay chassis, but that was something I had to insist on. So each sled of 12 drives is 9 data drives, 2 parity drives and 1 hot spare. The Integrator thought I was nuts but we insisted.

We set up our previous system in a similar fashion, using Infortrend 24-bay RAIDs. Each chassis had 2 RAID groups of 10+2 and 9+2 with a global hot spare for each chassis. The system before it had 16-bay Arena RAIDs set up as RAID5 in a single 14+1 RAID Group per chassis with one hot spare. We lost data at least 4 times in 3 years so when we replaced the hardware in 2006, we set the new systems up as described. In the 7 years we ran with that configuration, we never lost any data.

Quintin,

To answer the basic how to calculate any RAID type you need to determine the 'usable' space which is post OS formatted space.

An online calculator that I have our sales folks use is here... Usable Space RAID calc.

Use Carl's excellent advice to determine how to physically build up the overall RAID solution. Sure it is more expensive up front..BUT... how expensive will it be when a less optimal solution is implemented and the data gets lost? This is the question the end customer must answer for you... but be sure to strongly suggest a better solution.

what....? RAID5 on setups this big? If you manage to recover the RAID without a problem nor a sweat I'll buy a lotery ticket..!

RAID10 is you're way to go here.

Fast rebuilds, super realiabe. OBR10

One Big RAID 10: The new standard in server storage - Spiceworks

Quick comment. RAID 5/6 does NOT store parity data on the "extra" drive. The parity is distributed across all drives. When calculating space,, you subtract 1x or 2x drives, but the parity is not solely on that drive.

I also like to recommend that an extra drive (or more) be purchased and kept on the shelf in the server room, so that if there is a failure, it can be replaced ASAP, without having to worry about finding the model number, trying to procure it, and install it in time.

There is a need to be mindful of the difference between drive manufacturer decimal based TB calcs based on 1,000,000 bytes per megebyte & historic binary megabyte calculations based on 1,048,576 bytes per megabyte. So a 4TB decimal based drive actually stores only 3.64TB of binary calculated data.

When calculating the storage needs, just need to be aware of whether the total storage requirement is a decimal or binary based figure. Binary means that an extra 9.05% of storage is needed to cover the difference.

So a 4TB decimal based drive actually stores only 3.64TB of binary calculated data.

To be clear it does still store 4,000,000,000,000 bytes.

But the when the OS reports on it using units of KB, MB, GB or TB it will seem less since 1K (as the OS sees it) = 1024, and a MB = 1024x1024 and so on. But these are technically KiB, MiB, GiB, TiB.

On the other hand network bandwidth is reported using plain old Mb/sec, which is 1,000,000 bits per second. So when doing these types of capacity calculations, where you are buying new drives, you should not need to convert between the two types of TBs. Just stick with all zeroes.

When the OS tells you that you 10 TB left though, that's when you need to convert from binary to compare correctly against your bitrate.

Binary means that an extra 9.05% of storage is needed to cover the difference.

This is only true for TB, each unit is off by a increasing percentage. I would instead recommend using a calculator like Mike D. links to above.